## Question

Given the `root`

of a binary tree, invert the tree, and return *its root*.

**Example 1:**

```
Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
```

## Algorithm

This problem is obvious, for such problem(handle entire tree), we could keep eyes on single node action, and implement it with recursion.

For this question, What a node do is that it's left and right node change position, so as their children. So we do a swap for each node's left and right, and do such thing for the new left and right recursively until the node is null and has no child.

## Code

This question is a little vague it didn't mention if we should create a new tree a do the invert in-place. So I have two implementation.

No in place:

/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public TreeNode invertTree(TreeNode root) { if (root == null) return root; TreeNode newRoot = new TreeNode(root.val); helper(root, newRoot); return newRoot; } private void helper(TreeNode root, TreeNode newRoot) { if (root == null) return; if (root.right != null) newRoot.left = new TreeNode(root.right.val); if (root.left != null) newRoot.right = new TreeNode(root.left.val); helper(root.left, newRoot.right); helper(root.right, newRoot.left); } }

**In place**:

class Solution { public TreeNode invertTree(TreeNode root) { if (root == null) return root; TreeNode temp = root.left; root.left = root.right; root.right = temp; invertTree(root.left); invertTree(root.right); return root; } }